Question: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $z \neq 0$. $a = \dfrac{20z + 90}{6z} \times \dfrac{2}{10z + 45} $
Explanation: When multiplying fractions, we multiply the numerators and the denominators. $a = \dfrac{ (20z + 90) \times 2 } { 6z \times (10z + 45) } $ $ a = \dfrac {2 \times 10(2z + 9)} {6z \times 5(2z + 9)} $ $ a = \dfrac{20(2z + 9)}{30z(2z + 9)} $ We can cancel the $2z + 9$ so long as $2z + 9 \neq 0$ Therefore $z \neq -\dfrac{9}{2}$ $a = \dfrac{20 \cancel{(2z + 9})}{30z \cancel{(2z + 9)}} = \dfrac{20}{30z} = \dfrac{2}{3z} $